Solving Trigonometry Problems

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\(\displaystyle \beginx \frac=\frac\,\,\,\,\,\,\,\,\,\,\,\,x \frac=\frac\\\,\,\,x=-\frac\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\frac\end\) Note that (by looking at Unit Circle) this can be simplified to \(\displaystyle \\) (Remember that you add \(\pi k\) instead of \(2\pi k\) for tan and cot).

\(\displaystyle \begin\frac=\sin \theta \cos \theta \\\cos \theta =4\theta \cos \theta \\4\theta \cos \theta -\cos \theta =0\\\cos \theta \left( \right)=0\\\cos \theta \left( \right)\left( \right)=0\\\cos \theta =0\text\,\,\text\,\,\sin \theta =\pm \frac\,\\\\\\end\) Note that you can check these in a graphing calculator (radian mode) by putting the left-hand side of the equation into \(\) and the right-hand side into \(\) and get the intersection.

Note that \(k\) represents all integers \(\left( k\in \mathbb \right)\).

Note also that I’m using “fancy” notation; you may not be required to do this., depending on your book or teacher.

You won’t get the exact answers, but you can still compare to the exact answers you got above. This is because we could have fewer or more solutions in the Unit Circle, and thus for all real solutions when we add the \(2\pi k\) or \(\pi k\).

So when we solve these types of trig problems, we always want to solve for the General Solution first (even if we’re asked to get the solutions between solutions.Let’s start out with solving fairly simple Trig Equations and getting the solutions from \(\left[ 0,2\pi \right)\), or \(\left[ \right)\).Here is the Unit Circle again so we can “pick off” the answers from it: Notice how sometimes we have to divide up the equation into two separate equations, like when the argument of the trig function is an expression, like \(\displaystyle \theta \frac\).If we want \(\displaystyle \left( \right)\) for example, like in the The Inverse Trigonometric Functions section, we only pick the answers from Quadrants I and IV, so we get \(\displaystyle \frac\) only.But if we are solving \(\displaystyle \sin \left( x \right)=\frac\) we get \(\displaystyle \frac\) and \(\displaystyle \frac\) in the interval \(\); there are no domain restrictions.As another example, for \(\cos \left( \frac \right)\), we’ll only get one solution instead of the normal two.Let’s do some problems, finding the general solutions first, and then finding the solutions in the interval.We learned how to factor Quadratic Equations in the Solving Quadratics by Factoring and Completing the Square section.Note that when we factor trig equations to find solutions, like we do with “regular” equations, we never just divide a factor out from each side.\(\displaystyle \begin\tan \left( \right)=1\\frac \frac=\frac\,\,\,\,\,\,\,\,\,\,\frac \frac=\frac\\,\,\,\frac=-\frac\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac=\frac\\,\,\,\theta =-\frac\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta =\frac\end\) Since trig functions go on and on in both directions of the \(x\)-axis, we’ll also have to know how to solve trig equations over the set of real numbers; this is called finding the general solutions for these equations.We still use the Unit Circle to do this, but we have to think about adding and subtracting multiples of \(2\pi \) for the sin, cos, csc, and sec functions (since \(2\pi \) is the period for them), and \(\pi \)for the tan and cot functions (since \(\pi \) is the period for them).

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